Question: Let $g(x)=\dfrac{\sqrt{2x}}{\ln(x)}$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{\sqrt{2x}}-\dfrac1x$ (Choice B) B $\dfrac{1}{[\ln(x)]^2}\left(\dfrac{\ln(x)}{\sqrt{2x}}-\dfrac{\sqrt{2x}}{x}\right)$ (Choice C) C $\dfrac{1}{[\ln(x)]^2}\left(\dfrac{1}{\sqrt{2x}}-\dfrac1x\right)$ (Choice D) D $\dfrac{x}{\sqrt{2x}}$
$g$ is a quotient of a composite function and another function. Let... $u(x)=\sqrt{x}$ $v(x)=2x$ $w(x)=\ln(x)$... then $g(x)=\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}$. To find $g'(x)$, we will need to use the quotient rule and the chain rule! $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left[\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}\left[u\Bigl(v(x)\Bigr)\right]w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Quotient rule}} \\\\ &=\dfrac{u'\Bigl(v(x)\Bigr)v'(x)w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=\dfrac{1}{2\sqrt{x}}$ $v'(x)=2$ $w'(x)=\dfrac1x$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}\dfrac{{u'\Bigl(v(x)\Bigr)}v'(x)w(x)-{u\Bigl(v(x)\Bigr)}w'(x)}{[w(x)]^2} \\\\ &=\dfrac{{\dfrac{1}{2\sqrt{2x}}}(2)\left(\ln(x)\right)-{\sqrt{2x}}\left(\dfrac1x\right)}{(\ln(x))^2} \\\\ &=\dfrac{\dfrac{\ln(x)}{\sqrt{2x}}-\dfrac{\sqrt{2x}}{x}}{[\ln(x)]^2} \\\\ &=\dfrac{1}{[\ln(x)]^2}\left(\dfrac{\ln(x)}{\sqrt{2x}}-\dfrac{\sqrt{2x}}{x}\right) \end{aligned}$ In conclusion, $g'(x)=\dfrac{1}{[\ln(x)]^2}\left(\dfrac{\ln(x)}{\sqrt{2x}}-\dfrac{\sqrt{2x}}{x}\right)$.